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Latent Heat

100 g of icel...

Question

100 g of ice\lbrack latent heat 80 cal/g at 0 C\rbrack is mixed with 100 g of water at 80 C\lbrack specific heat 1 cal/g\rbrack.The final temperature of the mixture will be ?

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100 g

of ice (latent heat

80 {cal g}^{- 1}

0^{\circ} C

100 g

of water (specific heat

1 {cal g}^{- 1} /^{\circ} C

80^{\circ} C

. The final temperature of the mixture will be

100 g

of ice (latent heat

80 {cal g}^{- 1}

0^{\circ} C

100 g

of water (specific heat

1 {cal g}^{- 1} /^{\circ} C

80^{\circ} C

. The final temperature of the mixture will be

100 g

(latent heat 80 {cal g}^{- 1})

0^{\circ} C

100 g

of water (specific heat

1 {cal g}^{- 1 \circ} C^{- 1}

80^{\circ} C

. The final temperature of the mixture will be:

100 g

of ice (latent heat

80 c a l g^{- 1}

0^{o} C

) is mixed with

100 g

of water (specific heat 1 cal

g^{- 1} a t^{o} C^{- 1}

80^{o} C

. The final temperature of the mixture will be:

1 g

100^{\circ} C

and an equal mass of ice at

0^{\circ} C

are mixed. The temperature of the mixture in steady state will be : (latent heat of steam

= 540 c a l / g

, latent heat of ice

= 80 c a l / g

, specific heat of water

= 1 c a l / g^{\circ} C

)

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